140 XIII INTEGRATION (1 3.9.1 3) In order that a mapping/ : X -* R should be integrable, it is necessary and sufficient that f should be measurable and that the upper integral ^*(|/|) be finite. Necessity. Since |/| is integrable (13.7.4), clearly ju*(|/|) is finite. Next we remark that, for each integer n, there exists by hypothesis a function/, e > such that/rg/, and X/« -/) £ V« (13.7.1). Replacing/, by inf / (12.7.5), we may assume that the sequence (/,) is decreasing. If g is the limit of this sequence (equal to its lower envelope), we have f^g and ^(g—f) = lim n(fn —f) — 0 (13.8.1). In other words, /is almost everywhere equal to n-xx) the limit of the sequence (/,). Now all functions belonging to J are measurable (1.3.9.11), and another application of Egoroff's theorem (13.9.10) then proves that /is measurable. Sufficiency. We may assume that/is finite, by replacing/by an equivalent function (13.6.4). By (13.9.12) there exists a sequence^) of measurable step functions with compact supports such that \gn\^ \f\ and the sequence (gn) converges to /almost everywhere. Now gn is a linear combination of charac- teristic functions of relatively compact measurable sets ; since such sets are integrable (13.9.2), it follows that gn is integrable. Hence, applying (13.8.4) again, /is integrable. (13.9.14) (i) If f is an integrable function and A is a measurable set, then the function fq>^ is integrable. (ii) If(An) is an increasing sequence of integrable sets, whose union is the complement of a negligible set N, then \fdfjL = lim \f<pAndn. J »-*«> J We may assume without loss of generality that / is everywhere finite (13.6.4). Then/(pA is measurable (13.9.6), and we have from which (i) follows by (13.9.13). The assertion (ii) follows from the dominated convergence theorem (13.8.4). In the situation of (13.9.14), we put j fyA dp = JA /<//*, or JA/(*) dfi(x); this number is called the integral off over A. Also (if A is measurable) we write J * fdu in place of J *fq>A dp.ned from semi-continuous